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Java streams map class property list to flattened map

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3

How can we turn a List<Foo> towards a Map<propertyA, List<propertyB>> in the most optimal way by using java streams.

Beware: propertyA is NOT unique

//pseudo-code
class Foo
 propertyA //not unique
 List<propertyB>

So far I have the following:

fooList.stream()
 .collect(Collectors.groupingBy(Foo::propertyA, 
 Collectors.mapping(Foo::propertyB, Collectors.toList())))

Resulting into a Map<propretyA, List<List<propretyB>>> which is not yet flattened for its value.

java java-stream

share|improve this question

edited Mar 13 at 19:44

Ruslan

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asked Mar 13 at 19:28

TerenceTerence

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add a comment | 

3

How can we turn a List<Foo> towards a Map<propertyA, List<propertyB>> in the most optimal way by using java streams.

Beware: propertyA is NOT unique

//pseudo-code
class Foo
 propertyA //not unique
 List<propertyB>

So far I have the following:

fooList.stream()
 .collect(Collectors.groupingBy(Foo::propertyA, 
 Collectors.mapping(Foo::propertyB, Collectors.toList())))

Resulting into a Map<propretyA, List<List<propretyB>>> which is not yet flattened for its value.

java java-stream

share|improve this question

edited Mar 13 at 19:44

Ruslan

5,04111 gold badge1313 silver badges3131 bronze badges

asked Mar 13 at 19:28

TerenceTerence

7,76111 gold badge1010 silver badges1919 bronze badges

add a comment | 

How can we turn a List<Foo> towards a Map<propertyA, List<propertyB>> in the most optimal way by using java streams.

Beware: propertyA is NOT unique

//pseudo-code
class Foo
 propertyA //not unique
 List<propertyB>

So far I have the following:

fooList.stream()
 .collect(Collectors.groupingBy(Foo::propertyA, 
 Collectors.mapping(Foo::propertyB, Collectors.toList())))

Resulting into a Map<propretyA, List<List<propretyB>>> which is not yet flattened for its value.

java java-stream

share|improve this question

edited Mar 13 at 19:44

Ruslan

5,04111 gold badge1313 silver badges3131 bronze badges

asked Mar 13 at 19:28

TerenceTerence

7,76111 gold badge1010 silver badges1919 bronze badges

//pseudo-code
class Foo
 propertyA //not unique
 List<propertyB>

fooList.stream()
 .collect(Collectors.groupingBy(Foo::propertyA, 
 Collectors.mapping(Foo::propertyB, Collectors.toList())))

share|improve this question

edited Mar 13 at 19:44

Ruslan

5,04111 gold badge1313 silver badges3131 bronze badges

asked Mar 13 at 19:28

TerenceTerence

7,76111 gold badge1010 silver badges1919 bronze badges

share|improve this question

edited Mar 13 at 19:44

Ruslan

5,04111 gold badge1313 silver badges3131 bronze badges

asked Mar 13 at 19:28

TerenceTerence

7,76111 gold badge1010 silver badges1919 bronze badges

edited Mar 13 at 19:44

Ruslan

5,04111 gold badge1313 silver badges3131 bronze badges

edited Mar 13 at 19:44

Ruslan

5,04111 gold badge1313 silver badges3131 bronze badges

asked Mar 13 at 19:28

TerenceTerence

7,76111 gold badge1010 silver badges1919 bronze badges

asked Mar 13 at 19:28

TerenceTerence

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3 Answers
3

active

oldest

votes

5

You could use Java 9+ Collectors.flatMapping:

Map<propretyA, List<propretyB>> result = fooList.stream()
 .collect(Collectors.groupingBy(Foo::propertyA, 
 Collectors.flatMapping(foo -> foo.propertyB().stream(), 
 Collectors.toList())));

Another way (Java8+) is by using Collectors.toMap as in this answer.

And yet another Java8+ way is to just not use streams, but use Map.computeIfAbsent instead:

Map<propretyA, List<propretyB>> result = new LinkedHashMap<>();
fooList.forEach(foo -> result.computeIfAbsent(foo.propertyA(), k -> new ArrayList<>())
 .addAll(foo.propertyB()));

share|improve this answer

edited Mar 13 at 20:35

answered Mar 13 at 20:28

Federico Peralta SchaffnerFederico Peralta Schaffner

24.5k55 gold badges3838 silver badges8383 bronze badges

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  …and if Java 9 is not an option, a backport of flatMapping is at the end of this answer
  
  – Holger
  Mar 13 at 22:19
  

  
  
  
  

add a comment | 

4

This should do it:

 fooList.stream()
 .collect(Collectors.toMap(Foo::getPropertyA, Foo::getPropertyBList, (l1, l2) -> 
 l1.addAll(l2);
 return l1;
 ));

share|improve this answer

answered Mar 13 at 20:07

Not a JDNot a JD

1,40211 silver badge1212 bronze badges

add a comment | 

1

You could use Collectors.toMap with merge function where you have to decide which List<propertyB> has to be used when duplicating keys:

Map<propertyA, List<propertyB>> collect = list.stream()
 .collect(Collectors.toMap(Foo::getPropertyA, Foo::getList, (l1, l2) -> ???));

Otherwise you can get IllegalStateException. From javadoc:

If the mapped keys contains duplicates (according to Object.equals(Object)), an IllegalStateException is thrown when the collection operation is performed.

It is up to you how to implement merge function, but usually you just want to return the union of 2 lists like: l1.addAll(l2); return l1;

share|improve this answer

edited Mar 26 at 9:14

answered Mar 13 at 19:36

RuslanRuslan

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add a comment | 

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5

You could use Java 9+ Collectors.flatMapping:

Map<propretyA, List<propretyB>> result = fooList.stream()
 .collect(Collectors.groupingBy(Foo::propertyA, 
 Collectors.flatMapping(foo -> foo.propertyB().stream(), 
 Collectors.toList())));

Another way (Java8+) is by using Collectors.toMap as in this answer.

And yet another Java8+ way is to just not use streams, but use Map.computeIfAbsent instead:

Map<propretyA, List<propretyB>> result = new LinkedHashMap<>();
fooList.forEach(foo -> result.computeIfAbsent(foo.propertyA(), k -> new ArrayList<>())
 .addAll(foo.propertyB()));

share|improve this answer

edited Mar 13 at 20:35

answered Mar 13 at 20:28

Federico Peralta SchaffnerFederico Peralta Schaffner

24.5k55 gold badges3838 silver badges8383 bronze badges

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  …and if Java 9 is not an option, a backport of flatMapping is at the end of this answer
  
  – Holger
  Mar 13 at 22:19
  

  
  
  
  

add a comment | 

5

You could use Java 9+ Collectors.flatMapping:

Map<propretyA, List<propretyB>> result = fooList.stream()
 .collect(Collectors.groupingBy(Foo::propertyA, 
 Collectors.flatMapping(foo -> foo.propertyB().stream(), 
 Collectors.toList())));

Another way (Java8+) is by using Collectors.toMap as in this answer.

And yet another Java8+ way is to just not use streams, but use Map.computeIfAbsent instead:

Map<propretyA, List<propretyB>> result = new LinkedHashMap<>();
fooList.forEach(foo -> result.computeIfAbsent(foo.propertyA(), k -> new ArrayList<>())
 .addAll(foo.propertyB()));

share|improve this answer

edited Mar 13 at 20:35

answered Mar 13 at 20:28

Federico Peralta SchaffnerFederico Peralta Schaffner

24.5k55 gold badges3838 silver badges8383 bronze badges

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  …and if Java 9 is not an option, a backport of flatMapping is at the end of this answer
  
  – Holger
  Mar 13 at 22:19
  

  
  
  
  

add a comment | 

You could use Java 9+ Collectors.flatMapping:

Map<propretyA, List<propretyB>> result = fooList.stream()
 .collect(Collectors.groupingBy(Foo::propertyA, 
 Collectors.flatMapping(foo -> foo.propertyB().stream(), 
 Collectors.toList())));

Another way (Java8+) is by using Collectors.toMap as in this answer.

And yet another Java8+ way is to just not use streams, but use Map.computeIfAbsent instead:

Map<propretyA, List<propretyB>> result = new LinkedHashMap<>();
fooList.forEach(foo -> result.computeIfAbsent(foo.propertyA(), k -> new ArrayList<>())
 .addAll(foo.propertyB()));

share|improve this answer

edited Mar 13 at 20:35

answered Mar 13 at 20:28

Federico Peralta SchaffnerFederico Peralta Schaffner

24.5k55 gold badges3838 silver badges8383 bronze badges

Map<propretyA, List<propretyB>> result = fooList.stream()
 .collect(Collectors.groupingBy(Foo::propertyA, 
 Collectors.flatMapping(foo -> foo.propertyB().stream(), 
 Collectors.toList())));

Map<propretyA, List<propretyB>> result = new LinkedHashMap<>();
fooList.forEach(foo -> result.computeIfAbsent(foo.propertyA(), k -> new ArrayList<>())
 .addAll(foo.propertyB()));

share|improve this answer

edited Mar 13 at 20:35

answered Mar 13 at 20:28

Federico Peralta SchaffnerFederico Peralta Schaffner

24.5k55 gold badges3838 silver badges8383 bronze badges

answered Mar 13 at 20:28

Federico Peralta SchaffnerFederico Peralta Schaffner

24.5k55 gold badges3838 silver badges8383 bronze badges

answered Mar 13 at 20:28

Federico Peralta SchaffnerFederico Peralta Schaffner

24.5k55 gold badges3838 silver badges8383 bronze badges

4

This should do it:

 fooList.stream()
 .collect(Collectors.toMap(Foo::getPropertyA, Foo::getPropertyBList, (l1, l2) -> 
 l1.addAll(l2);
 return l1;
 ));

share|improve this answer

answered Mar 13 at 20:07

Not a JDNot a JD

1,40211 silver badge1212 bronze badges

add a comment | 

4

This should do it:

 fooList.stream()
 .collect(Collectors.toMap(Foo::getPropertyA, Foo::getPropertyBList, (l1, l2) -> 
 l1.addAll(l2);
 return l1;
 ));

share|improve this answer

answered Mar 13 at 20:07

Not a JDNot a JD

1,40211 silver badge1212 bronze badges

add a comment | 

This should do it:

 fooList.stream()
 .collect(Collectors.toMap(Foo::getPropertyA, Foo::getPropertyBList, (l1, l2) -> 
 l1.addAll(l2);
 return l1;
 ));

share|improve this answer

answered Mar 13 at 20:07

Not a JDNot a JD

1,40211 silver badge1212 bronze badges

 fooList.stream()
 .collect(Collectors.toMap(Foo::getPropertyA, Foo::getPropertyBList, (l1, l2) -> 
 l1.addAll(l2);
 return l1;
 ));

share|improve this answer

answered Mar 13 at 20:07

Not a JDNot a JD

1,40211 silver badge1212 bronze badges

answered Mar 13 at 20:07

Not a JDNot a JD

1,40211 silver badge1212 bronze badges

answered Mar 13 at 20:07

Not a JDNot a JD

1,40211 silver badge1212 bronze badges

1

You could use Collectors.toMap with merge function where you have to decide which List<propertyB> has to be used when duplicating keys:

Map<propertyA, List<propertyB>> collect = list.stream()
 .collect(Collectors.toMap(Foo::getPropertyA, Foo::getList, (l1, l2) -> ???));

Otherwise you can get IllegalStateException. From javadoc:

If the mapped keys contains duplicates (according to Object.equals(Object)), an IllegalStateException is thrown when the collection operation is performed.

It is up to you how to implement merge function, but usually you just want to return the union of 2 lists like: l1.addAll(l2); return l1;

share|improve this answer

edited Mar 26 at 9:14

answered Mar 13 at 19:36

RuslanRuslan

5,04111 gold badge1313 silver badges3131 bronze badges

add a comment | 

1

You could use Collectors.toMap with merge function where you have to decide which List<propertyB> has to be used when duplicating keys:

Map<propertyA, List<propertyB>> collect = list.stream()
 .collect(Collectors.toMap(Foo::getPropertyA, Foo::getList, (l1, l2) -> ???));

Otherwise you can get IllegalStateException. From javadoc:

If the mapped keys contains duplicates (according to Object.equals(Object)), an IllegalStateException is thrown when the collection operation is performed.

It is up to you how to implement merge function, but usually you just want to return the union of 2 lists like: l1.addAll(l2); return l1;

share|improve this answer

edited Mar 26 at 9:14

answered Mar 13 at 19:36

RuslanRuslan

5,04111 gold badge1313 silver badges3131 bronze badges

add a comment | 

You could use Collectors.toMap with merge function where you have to decide which List<propertyB> has to be used when duplicating keys:

Map<propertyA, List<propertyB>> collect = list.stream()
 .collect(Collectors.toMap(Foo::getPropertyA, Foo::getList, (l1, l2) -> ???));

Otherwise you can get IllegalStateException. From javadoc:

If the mapped keys contains duplicates (according to Object.equals(Object)), an IllegalStateException is thrown when the collection operation is performed.

It is up to you how to implement merge function, but usually you just want to return the union of 2 lists like: l1.addAll(l2); return l1;

share|improve this answer

edited Mar 26 at 9:14

answered Mar 13 at 19:36

RuslanRuslan

5,04111 gold badge1313 silver badges3131 bronze badges

Map<propertyA, List<propertyB>> collect = list.stream()
 .collect(Collectors.toMap(Foo::getPropertyA, Foo::getList, (l1, l2) -> ???));

share|improve this answer

edited Mar 26 at 9:14

answered Mar 13 at 19:36

RuslanRuslan

5,04111 gold badge1313 silver badges3131 bronze badges

answered Mar 13 at 19:36

RuslanRuslan

5,04111 gold badge1313 silver badges3131 bronze badges

answered Mar 13 at 19:36

RuslanRuslan

5,04111 gold badge1313 silver badges3131 bronze badges